1) Let's choose arbitrary and assume that . , so there exists an injection . so . There for function is also injection. That means that , so . is a set of ordinals and it is transitive, so is ordinal.

Very good.

I would put it this way:

Suppose g in b in a. So there is an injection from b into A. Also, the identity function on g is an injection from g into b. So, by composition of functions, we have an injection from g into A. And g is an ordinal. So g in A. So a is epsilon-transitive. So, since a is an epsilon-transitive set of ordinals, we have that a is an ordinal.

Originally Posted by Ester

2) I have only ideas for this. I thought, I could begin like this:
Let's choose arbitrary . That means , so .
Counterexample: .

You've almost got it (except, I don't see why you use the word 'counterexample' there).

These are the relevant items now:

1. b is equinumerious with a
2. b is dominated by A
3. a is an ordinal
4. I don't want to tell you this one (but you know it) since telling you would give away the answer.
5. Another simple fact I don't want to give away.

So now just draw a contradiction from 1-5. Let me know what you come up with ...

You've almost got it (except, I don't see why you use the word 'counterexample' there).

Sorry, English is not my native language. I think the word what I was thinking was counterassumption (?).

Originally Posted by MoeBlee

These are the relevant items now:

1. b is equinumerious with a
2. b is dominated by A
3. a is an ordinal
4. I don't want to tell you this one (but you know it) since telling you would give away the answer.
5. Another simple fact I don't want to give away.

So now just draw a contradiction from 1-5. Let me know what you come up with ...

This is what I came up with:

and , so there for . is an ordinal, so . That means that , which is contradiction. There for the counterassumption is false and orginal statement is true.

Using the definition of cardinals, I get that card(A) is an ordinal and , so . That means that . , so .

Very good. A couple of notes:

(1) We have that card(A) is equinumerous with A by virtue of the numeration theorem (derived from the axiom of choice and the axiom schema of replacement). (Though, by the Fregean method, this particular proof will work even without numeration theorem.)

(2) Just remember that card(A) in a implies that card(A) is cardinal-less-than a by the fact that both card(A) and a are cardinals (as opposed to merely ordinals, since for certain ordinals we have k in j and k equinumerous with j).

Here's another version (basically the same as yours):

Suppose it is not the case that card(A) < a. So a is less than or equal card(A). (This does not require the axiom of choice, since we're not relying on domination trichotomy holding among all sets but rather only that cardinal less-than trichotomy holds among all cardinals.) So a is dominated by A. So a in a, which is a contradiction, since a is an ordinal (also contradicts axiom of regularity).

Originally Posted by Ester

Counterassumption: There exists a cardinal such that . We have , which means . So is an ordinal and . , so .
Now we have , which leads to . This is contradiction.

Very good. Note that the numeration theorem was used to infer "A is strictly dominated by k" from "card(A) is cardinal-less-than k".